Bourbaki

Exercise 7, § 4

Commutative Algebra, VIII, p. 90

a) Let l be an integer ≥1\geq 1 and n \in \mathbf{N}. Show that there exists a unique decreasing sequence of integers: al,l(n)≥al,l−1(n)≥⋯≥al,1(n)≥0a_{l,l}(n) \geq a_{l,l-1}(n) \geq \cdots \geq a_{l,1}(n) \geq 0 such that

n=∑i=1l(al,i(n)+i−1i). n = \sum_{i=1}^l \binom{a_{l,i}(n) + i - 1}{i}.

Show that one has n≤mn \leq m if and only if (al,l(n),al,l−1(n),...,al,1(n))(a_{l,l}(n), a_{l,l-1}(n), ..., a_{l,1}(n)) is less than (al,l(m),...,al,1(m))(a_{l,l}(m), ..., a_{l,1}(m)) for the lexicographic order. Set

∂l(n)=∑i=1l(al,i(n)+ii+1), \partial_l(n) = \sum_{i=1}^l \binom{a_{l,i}(n) + i}{i + 1},

and for l≥2l \geq 2, denote by ∂l∗(n)\partial_l^*(n) the smallest integer p such that ∂l−1(p)≥n\partial_{l-1}(p) \geq n. Calculate al+1,i(∂l(n)),al−1,i(∂l∗(n))a_{l+1,i}(\partial_l(n)), a_{l-1,i}(\partial_l^*(n)) in terms of the al,i(n)a_{l,i}(n). b) We say that a mapping H : \mathbf{N} \to \mathbf{N} is a Macaulay function if

{H(0)=1,H(l+1)≤∂l(H(l))for every l≥1. \begin{cases} H(0) = 1, \\ H(l + 1) \leq \partial_l(H(l)) \quad \text{for every } l \geq 1. \end{cases}

Show that it comes to the same thing as saying that

{H(0)=1,∂l∗H(l)≤H(l−1)for every l≥2. \begin{cases} H(0) = 1, \\ \partial_l^* H(l) \leq H(l - 1) \quad \text{for every } l \geq 2. \end{cases}

Let H:N→NH : \mathbf{N} \to \mathbf{N} be a Macaulay function. Show that only two cases are possible: α) There exists r∈Nr \in \mathbf{N} such that H(j)=0H(j) = 0 for all j≥rj \geq r. β) There exist r∈Nr \in \mathbf{N} and a finite sequence of integers

b0≥b1≥⋯≥br≥0 b_0 \geq b_1 \geq \cdots \geq b_r \geq 0
such that for all j≥rj \geq r, one has
(∗)H(j)=∑n=0r(bn−n+jbn). (*) \quad H(j) = \sum_{n=0}^r \binom{b_n - n + j}{b_n}.
Show that rr and the sequence b0≥b1≥⋯≥br≥0b_0 \geq b_1 \geq \cdots \geq b_r \geq 0 are uniquely determined by the function HH. We set d(H)=0d(H) = 0 in case α) and d(H)=b0+1d(H) = b_0 + 1 in case β), so that d(H)−1d(H) - 1 is the degree of the polynomial () in the variable jj. Let a(H)jd(H)−1(d(H)−1)!a(H) \frac{j^{d(H)-1}}{(d(H)-1)!} be the term of highest degree of the polynomial (). Show that a(H)a(H) is an integer >0> 0. Show that a polynomial in one variable jj with coefficients in Q\mathbf{Q}, which takes integer values on the integers and which is strictly positive for jj sufficiently large, is not necessarily of the form (*). c) Let H:N→NH : \mathbf{N} \to \mathbf{N} be a Macaulay function. We set
SH(T)=∑n=0∞H(n)Tn, S_H(T) = \sum_{n=0}^\infty H(n) T^n,
so that SH(T)∈Z[[T]]S_H(T) \in \mathbf{Z}[[T]]. Show that SH(T)∈Z[T]S_H(T) \in \mathbf{Z}[T] or else that there exist a finite sequence of integers
c0≥c1≥⋯≥cr>0 c_0 \geq c_1 \geq \cdots \geq c_r > 0
uniquely determined by HH, and a polynomial P∈Z[T]P \in \mathbf{Z}[T] such that one has
(∗∗)SH(T)=∑n=0rTn(1−T)cn+P. (**) \quad S_H(T) = \sum_{n=0}^r \frac{T^n}{(1-T)^{c_n}} + P.
Show that c0=d(H)c_0 = d(H) and that a(H)a(H) is the number of integers cic_i such that ci=d(H)c_i = d(H). The rational fraction ∑n=0rTn(1−T)cn\sum_{n=0}^r \frac{T^n}{(1-T)^{c_n}} is called the asymptotic expansion of HH. Let A1,...,AdA_1, ..., A_d be integers, with Ad≠0A_d \neq 0, such that the rational fraction
∑n=1dAn(1−T)n−SH(T) \sum_{n=1}^d \frac{A_n}{(1-T)^n} - S_H(T)
is a polynomial. Show that the sequence (An)1≤n≤d(A_n)_{1 \leq n \leq d} uniquely determines and is uniquely determined by the asymptotic expansion of HH. In particular one has d=d(H),Ad=a(H)d = d(H), A_d = a(H). Examine the case d(H)=2d(H) = 2 in greater detail. d) Let H:N→NH : \mathbf{N} \to \mathbf{N} be a Macaulay function. Show that the following conditions are equivalent: (i) ∂l∗H(l)=H(l−1)\partial_l^* H(l) = H(l-1) for all l≥2l \geq 2, (ii) HH is the smallest of the Macaulay functions having the same asymptotic expansion as HH. Such Macaulay functions are said to be extreme. Let c0≥c1≥⋯≥cr>0c_0 \geq c_1 \geq \cdots \geq c_r > 0 be a sequence of integers. Show that there exists one and only one extreme Macaulay function having ∑n=0rTn(1−T)cn\sum_{n=0}^r \frac{T^n}{(1-T)^{c_n}} as asymptotic expansion. Show that, if HH is an extreme Macaulay function, one has H(1)=d(H)H(1) = d(H) or else H(1)=d(H)+1H(1) = d(H) + 1. Deduce that for every Macaulay function HH, one has H(1)≥d(H)H(1) \geq d(H). e) Let H:N→NH : \mathbf{N} \to \mathbf{N} be a Macaulay function. Show that the following properties are equivalent: (i) one has SH(T)=1(1−T)d(H)S_H(T) = \frac{1}{(1-T)^{d(H)}},

(ii) H(1)=d(H)H(1) = d(H), (iii) ∂lH(l)=H(l+1)\partial_l H(l) = H(l + 1) for all l⩾1l \geqslant 1. In particular, every function such that H(1)=d(H)H(1) = d(H) is extreme.

Solution

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Machine translation of the English, by gpt-5.4. Not checked by a person.